Explanation

(b) Given that,
$\frac{l_{1}}{l_{2}} = \frac{4}{3}and\frac{r_{1}}{r_{2}} = \frac{2}{3}$
Here, $l_{1}$and $l_{2}$are the length of the wires while $r_{1}$and $r_{2}$ are the radii of the wires.
Now, we know that
$V = IR \Longrightarrow IR = constant$
$\Longrightarrow I_{1}R_{1} = I_{2}R_{2}$
$or\ \ \ \frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}}\ \ \ \ldots.(i)$
But we know that the resistance of the wire is
$\ \ \ \ \ \ \ \ \ R = \frac{\rho l}{A}$
Hence, from Eq (i)
$\frac{I_{1}}{I_{2}} = \frac{\rho l_{2}/A_{2}}{\rho l_{1}/A_{1}}$
Here,
$R_{1} = \frac{\rho l_{1}}{A_{1}},R_{2} = \frac{\rho l_{2}}{A_{2}}and\ \rho_{1} = \rho_{2} = \rho$
Because both wires are of same material.
$\therefore\ \ \frac{I_{1}}{I_{2}} = \frac{l_{2}A_{1}}{l_{1}A_{2}}$
$\Longrightarrow \frac{I_{1}}{I_{2}} = \frac{l_{2}\pi r_{1}^{2}}{l_{1}\pi r_{2}^{2}}\ $
$\Longrightarrow \frac{I_{1}}{I_{2}} = \frac{l_{2}r_{1}^{2}}{l_{2}r_{2}^{2}}$
Here, $\frac{l_{2}}{l_{1}} = \frac{3}{4}\ \ and\frac{r_{1}}{r_{2}} = \frac{2}{3}$
$\Longrightarrow \ \ \frac{I_{1}}{I_{2}} = \frac{3}{4} \times \left( \frac{2}{3} \right)^{2}$
Or $\frac{I_{1}}{I_{2}} = \frac{1}{3}$